Chapter 6: Force & Motion II. Lecture 13 9/23/09

Transcription

1 Chapter 6: Force & Motion II Lecture 13 9/23/09

2 The Drag Force & The Centripetal Force Goals for this Lecture: Describe the drag force between two objects. Relate the rag force to the terminal speed of falling objects Briefly review uniform circular motion Relate the centripetal force to uniform circular motion

3 Drag force is the non-solid equivalent of friction When do we deal with the drag force: Whenever an object moves in a fluid. A fluid can be a liquid (water, molasses,...) or a gas (air,...) Viscous fluid = more drag force Application point: At the surface of the moving objects Direction: The Drag Force: D Opposite to the direction of motion

4 Magnitude: C: Drag coefficient (Depends on the shape on an object) A: The effective cross section of the object (Depends on the shape on an object) Note: The Drag Force: D ρ: The density of the fluid v: The velocity of the object through the fluid If v = 0, D = 0 D = ½CρAv 2 To keep an object moving at a speed v requires a force F that is to v 2 (since F net = F - D = 0)

5 Example Auto fuel consumption vs velocity The frictional forces on a moving automobile consist of two major components a) a constant rolling resistance term and b) a velocity dependent drag force A Hummer H2 has the following properties F rolling = 300 N Drag coefficient: C = 0.57 (C Honda-Insight = 0.25) Area (looking from the front): A = 2.5 m 2 Density of air: ρ = 1.3 kg/m 3 How much more force must be supplied by the engine to keep the H2 moving at 100 mph, compared to 60 mph

6 Example Auto fuel consumption vs velocity 60 mph = 60 miles/hour*1600m/mile*1 hour/3600 s = 27 m/s F drag-60 = ½CρAv 2 = ½(0.57)(1.3 kg/m 3 )(2.5 m 2 )(27 m/s) 2 = 675 N F drag-100 = ½CρAv 2 = ½(0.57)(1.3 kg/m 3 )(2.5 m 2 )(45 m/s) 2 = 1876 N F tot-60 = 300 N N = 975 N F tot-100 = 300 N N = 2176 N F tot-100 / F tot-60 = 1976/775 = 2.23 < x worse fuel consumption

7 Terminal Speed Consider an object in free fall through air The forces acting on it are gravity and the drag force 1. The object has just been dropped v = 0, D = 0, F net,1 = F g - D = F g : object accelerates 2. The object has fell for a little time v > 0, D > 0, F net,2 = F g - D < F g : object still accelerates, but at a slower rate 3. The object has fell for more time v >> 0, D = F g, F net,2 = F g - D = 0 : object no longer accelerates, but falls at a constant speed D F g

8 Terminal Speed When the object has reached terminal speed D = F g --> 1/2 CρAv 2 = mgmt. Terminal speed depends on: Shape of the object (C, A) Size of the object (A) Mass of the object (m) Density of the air (ρ) v term = D 2mg CρA F g

9 Terminal Speeds Object Mass Area Terminal Speed m/s mph Skydiver 75 kg 0.7 m Baseball 145 g 42 cm Golf ball 46 g 14 cm Hail stone 0.48 g 0.79 cm Raindrop g 0.13 cm

10 Uniform Circular Motion An object moving with uniform circular motion has: Constant angular velocity: ω Constant speed (i.e. magnitude of velocity): v = ωr Changing velocity vector Acceleration towards the center (centripetal acceleration): a c = v 2 /r = ω 2 R Constant magnitude of acceleration Changing acceleration vector C

11 Centripetal Force: F c When do we deal with the centripetal force: Whenever an object moves in a curved trajectory Direction: Points towards the center of curvature, and always perpendicular to the velocity vector Magnitude: Newton 2nd law: F c = ma c --> F = mv 2 /R = mω 2 R Note: The centripetal force in the NET force perpendicular to the direction of motion F c C F c F c

12 The Physical Cause of Centripetal Forces Any force can be a centripetal force As long as it is perpendicular to the velocity

13 The Physical Cause of Centripetal Forces Any force can be a centripetal force Gravity: F g F g

14 The Physical Cause of Centripetal Forces Any force can be a centripetal force Friction: f x C C

15 The Physical Cause of Centripetal Forces Any force can be a centripetal force Tension: T T

16 The Physical Cause of Centripetal Forces Any force can be a centripetal force Normal force: N y C x

17 Example Arched bridge vs. Suspended bridge A car with mass m is driving at constant speed v across a bridge with radius of curvature R. What is the force which the car exerts on the bridge? Hint: The force which the car exerts on the bridge is equal to (in magnitude) to the force which the bridge exerts on the car (i.e. The normal force)

18 Example y x Arched bridge vs. Suspended bridge: N F g F N + F g = ma F N - F g = -ma c F N = F g - ma c = m(g - v 2 /R) N F g F N + F g = ma F N - F g = +ma c F N = F g + ma c = m(g + v 2 /R) A suspension bridge feels more stress from passing traffic

19 Example San Francisco car chase: How fast does a car have to moving in order to fly off the top of a curved road / hill with a radius of curvature of R meters?

20 Example San Francisco car chase: y N F N + F g = ma x F N - F g = -ma c 0 - mg = -ma c F g ma c = mg a c = g = v 2 /R v = Ra c